A spherical liquid drop of radius `R` is divided into eight the surface tension is `T`, then the work equal droplets. If the surface tension is `T`, then the work done in this process will be

Options

A. `2piR^(2)T`
B. `3piR^(2)T`
C. `4piR^(2)T`
D. `2piRT^(2)`

Answer

C

Explanation

Radius of the larger drop`=R`
Suppose radius of the droplets `=r`
Since volume will be remain constant
`4/3piR^(3)=8xx4/3pir^(2)` (as no. of droplets `=8`)
`:. R=((R^(3))/8)^(1/3)=R/2`
Therefore, work done `=` Increase in surface area `xx` Surface tension
`=[84pi(R/2)^(2)-4piR^(2)]xxT`
`=(8piR^(2)-4piR^(2))xxT=4piR^(2)T`